I usually do it like this: e.g. for 32 x 34: double one side and half the other side: 32 x 34 = 16 x 68 = 8 x 136 = 4 x 272 = 2 x 544 = 1088 It's easy to do this very quick in the head.
But any number that is not a product of two to a power can't be halved until it reaches 1, you eventually get fractions which are much harder to deal with and end in not one
It works with any prime number e.g.: 49 x 43 = 7 x 301 = 2107 27 x 61 = 3 x 549 = 1660 - 3 = 1647 This method is not always faster than the "common method", but in many cases it is.
Hey techmath, The method you've used might get tedious for 3 digit numbers. I've got a much easier way: 1) For 2 digit number: NOTE: we are dealing with 2 digit number, so we must have only one digit entries. Let's take (32)^2. -> Square each digit and write 'em down at both ends. Like--> 9____4. (3^2=9, 2^2=4). -> Now need to find the middle portion of answer, for that just multiply both digit and double the result, That implies, (3x2)x2 = 12. -> So for that, write 2 in blank space and carry 1 to next digit. Why? Refer the NOTE above! That is, 924 and then add 1 to 9. So we get -> 1024. 2) For 3 digit number: Let take (409)^2. NOTE: we are dealing with 3 digit number, so we must have only 2 digit entries and group the number in pair of two from right. -> Group the numbers from right (in pair of two) , so we have two groups as: Group 1:- 4, Group 2:- 09. Now the rest of the procedure is same. -> Follow same procedure. square the two groups and write at both ends. Like--> 16____81. (4^2=16, 9^2=81). -> Now need to find the middle portion of answer, for that just multiply both group and double the result, That implies, (4x9)x2 = 72. -> So answer is, 16 72 81 = 167281. 3) Another 3 digit number: Let's try with a tougher number:- (825)^2. -> Group the numbers from right (in pair of two) , so we have two groups as: Group 1:- 8, Group 2:- 25. Now the rest of the procedure is same. -> Follow same procedure. square the two groups and write at both ends. Like--> 64____25. Why? Refer NOTE for 3 digit numbers. We must have only 2 digit entries for 3 digit number. And (25)^2=625. So write 25 and take 6 as carry.---------------------------------------> (1). -> Now need to find the middle portion of answer, for that just multiply both group and double the result, That implies, (8x25)x2 = 400. But write only 00 and take 4 as carry----------------------------------> (2). Hence, Result will be:- 64_____25 --> 64 00 25, But we are yet to add the carries from equation (1) and (2). Now from (1) add carry 6 to 00. And from (2) add carry 4 to 64. Final answer: 68 06 25. P.S If anyone has any doubt, ask me! I'll try to clarify! Cheers!
Hey Lon Dong! :) Good doubt, I was expecting this somewhere down the line! I'm glad you asked. Follow same procedure! -> 9^2=81 Remember, the NOTE given for 2 digit no, only one digit will be written initially. Hence, ->Write as 1____1 (8, being taken as carry from both ends). -> Now (9x9)x2=162. Here lies the difference, since we got 3 digit result. Follow carefully from now. I'll rewrite the result, this time the value to be carried in brackets, just to help you understand better! (8) (8) -> 1___1 (Again, Remember only one digit will come in the blank space as we dealing with 2digit number). -> 162+8=170. Put 0 in the blank space and take 1,7 as carry. (1) (8) (7) -> 1 0 1. Now, it's again simple. Just add the carry to the corresponding digit below! -> 1+7=8. (1) (8) -> 8 0 1, and 1+8=9. -> 9 8 0 1. P.S: You'll find it difficult in the beginning, but believe me you'll be a master in this. Practice makes it perfect. I could do these calculations in 15-20 seconds and that's only because of practice. Just remember the NOTEs i've mentioned. Try for 999*999! Cheers!
I swear, this method is bloody brilliant!!! I'm squaring numbers in their hundreds! Just squared 280 in seconds in my head. After attempting to square numbers between 100 and 200, and finding that numbers such as 120 which end in a zero were easier to do by writing it as (12*10)^2=12^2 * 10^2=144*100=14,400, I found this method gets difficult after 200; and I then realised that ANY number between 100 and 1000 ending with a zero is EASY to square using tecmath's method (I actually JUST now figured out you can even for numbers above those in the 200's, as I just attempted to square 980!). I'm bloody amazed and excited man I swear ^^. You can still use this method for numbers in their hundreds which do not end in zero, however I have found they are quite difficult to do when the difference between the base you choose and the number is greater than 20, as memorising the square of numbers above 20 is infeasible, whilst those below are pretty easy. For example, choosing a base of 200, and doing 185^2: 200*170 + 15^2 = 34000 + 225 = 34225. If you aren't sure what I did, the difference between 185 and 200 is 15, so add and minus 15 from 185 and multiply those two together, and add the square of 15. If you still aren't sure then refer to tecmaths video above. Here it is easy to square 15, but if you try a number such as 167, the distance to 200 is 33, and I don't know about any of you, but I don't know 33^2 off the top of my head! In reflection though, squaring numbers even with a distance from the base greater than 20 seems to be a lot easier and faster than the traditional long multiplication method! I just attempted to square 167, having to square 32 doing the repeat process, and it still is a lot better than the alternative. Thank you tecmath for giving me the ability to square ANY/MOST numbers between 1 and 1000 in seconds/a-minute!
I like the duplex method. First, I will state a few duplex patterns and maybe you will notice a pattern. The duplex method can be used to square any number with any number of digits. D(a) = a^2 ( duplex of a single digit ) D(ab) = 2ab ( duplex of a two digit number ) D(abc) = 2ac + b^2 ( duplex of a three digit number ) D(abcd ) = 2ad + 2bc ( duplex of a four digit number ) D(abcde) = 2ae + 2bd + c^2 ( duplex of a five digit number ) so 33^2 = 3^2 I 2*3*3 I 3^2 33^2 = 9 I 18 I 9 perform any carries that are necessary and we get 1089
@@arcwand Any explanation should come with an example. I can't see any sense in Michael's version. But I can show how to square a number of any size using cross multiplying, and also by arcing. All methods used are variations of the squaring algorithm, but cross multiplying can use mixed numbers. The small example he showed doesn't appear to match his first outline, but is a well known vedic method. IE 137^2= 13^2/13x7x2/7^2= 169/182/49= 187/6/9 (with carryovers), =18,769 [the number of digits in the RHS separations depend on how many in the 2nd root. If it was 16^2=256, the RHS would use 2 digits as in 16, and so 56 would remain in that case. In the example it uses 7^2 or one digit]
0:36 O thanks! I want to thank my parents, my teachers and especially to my faithful friend, my calculator, because without it, this would not be possible.
The reason behind this is simple actually You can write the equation of first question like this :- (30+2)(34-2) Let 30 be 'a', 34 be 'b' and 2 be 'c' Now, (a+c)(b-c) Now ab+cb-ac-c^2 => ab + c(b-a) - c^2 => ab + c(2c) - c^2. (b-a = 2c) => ab + 2c^2 - c^2 = ab + c^2 = 30×34 + 2^2 = 1024. (Putting the values of a,b and c)
I have a different method which I thought of in high school which works really well for me. Let's take 32^2 again, then you find the nearest 10 just like you -> 30^2 (=900), then let's call (for the sake of the explanation) the difference between the 2 numbers: 'd' (=2 in this case). Then to get the answer you do: 900 + (30+32)*d = 900+62*2=1024. The great thing about this method is that if you want to calculate the square of broken numbers like 41.5^2, you can do the same thing: 1600 + (40+41.5)*1.5 = 1722.25. Or the square of 0.913? .81 + (.9+.913)*.013 = .81 + .01813 + .005439 = .833569! Okay, maybe that was a bit too hard...
both your and his method are built upon the same very basic maths: (a+b)^2 = a^2 + 2ab + b^2 The video lacks of that explanation though, and honestly i think you could as well just use the equation i gave
@@Player-hx1gs Not quite. The method used is actually conjugates, not square of a sum. In the video, (a+b)(a-b) = a² - b², and therefore (a+b)(a-b)+b² = a². However, Flik's method is slightly different: 'd' is defined as b-a, and multiplied with a+b.
I think it is also worth mentioning that either method works for numbers on either side of the rounding up line. Like 32^2 using the method of rounding up to the higher place of ten (40), and the difference being 8, 32-8=24. Which multiplying would be (40*20=800)+(40*4=160)=960. Adding the squared difference (8^2=64) 960+64=1024. On the other hand, using 77 and putting it to the nearest ten of 70 at a difference of 7, the multiplier would be 84. Thus it would (70*80=5600)+(70*4=280)=5880. Adding the squared difference (7^2=49), 5880+49=5929. ((Granted this is possible, I never said it was easier than your given on how to do this. I appreciate that this is out there in the first place.))
Thanks for the help. I am spending one month studying all these skills to try to improve in my calcula to no speed. This helps! Continue the good work.
The trick can be expressed by the equation x^2 = (x + a)(x - a) + a^2 Technically, we could choose any number "a" to add and subtract from x, then add back its square, and it would still work. But we choose "a" as the difference to the nearest 10 so that one of the (x +/- a) terms come out to round 10's, for easy mental multiplication.
For 32 squared I would do this: Picture it as 32 X 32 Units: 2 X 2 = 4 so you write down 4 Outer and inner: (2X3)+(2X3)=12 so you write down the 2 and carry the 1 Tens: 3 X 3 = 9 plus the carried 1 = 10 so you put down the 0 and carry the 1 Add the carried 1 from the end so you have 1024 It's the quickest way to do it in my opinion and you can do it in your head as long as you can visualise it and know your times tables solidly
You helped me so much. I'm in 8th grade and I have to do three one minute 10 problem quizzes of my squares and roots 1-25. Thanks so much! You earned yourself a subscriber!
This works. It is faster than doing it by hand most of the time. Of course, you have to be used to "storing" sub-calculations in your head while you "get the rest of it." That's the challenge for me, not so much the computations. There are a several methods that are almost instant: they are a little more sophisticated, BUT require less mental "storage" and computation. They are especially effective if you know your perfect squares up to 24 (not that hard to memorize). If you know those, you can square any number in your head REALLY fast with very little computation--like no more than two seconds. The hardest numbers are between 71 and 74, but they are still faster than the method in the video. PLEASE NOTE, I am not trying to discredit the video. It works and this guy has a bunch of great methods for doing all sorts of fun math computations that I hadn't seen and I think are wonderful. But for squares, there are four methods used for different ranges of numbers that are almost instant and less mentally taxing. The quickest of them is what you could call "base 50 on 25" (NOT the same base 50 method he uses in one of his multiplication videos). So, if someone asked you to square 57 (for example), you almost instantly blurt out "thirty two forty-nine" with almost no mental computation at all (only squaring the one's digit, 7, and adding 7 to 25). That forms in your head as 32, which you affix to 49. If asked 42 squared, you almost instantly say, "seventeen sixty-four" (slightly harder than the above number, but not much). The only thing you actually did computationally in your head is subtracted 8 from 25 (8 is the difference between 50 and 42), squared that difference, and affixed your results side-by-side. That forms in your head as 50-42 is 8. 25-8 is 17. 8 squared is 64. You now affix17 and 64. 1764. That last squared quantity will take the first two place values if it is a two digit number and three place values if it has three digits, meaning you will have to "overlap" a single digit when you affix the two quantities. So you always work with the distance between 50 and your number and "center" the result around 25: you either subtract it from 25 or add to 25. That result will start with the thousands place. You then square the distance from 50 affix it to what you had from the first step. Here's one more: 63 squared. It is "thirty-nine sixty-nine." To get it, you think of the difference between 50 and 63. So, 13. You add that thirteen to 25. So 38. This is actually 3800, but just think of it as 38; it's less to think about and "store" in your head. You now square 13. So 169 (you should have this memorized for these techniques). Now since the 169 is three digits long this time, you "overlap" the last digit of the 38 and the first digit of the 169. The "overlapping" is really addition. What you are doing is adding 3800 to 169. But don't even think of it that way. Only think of adding the overlapping digit: the 8 and the 1. So, you now think 39 rather than 38 and affix it to the last two digits of 169, so 69. You now have a 39 and a 69 joined to 3969. Again, don't even think of the actual big numbers or that you are adding them. You had three very simple calculations and an "affixing" or joining of the results. So you never really have to do any bigger computations, assuming you've memorized your squares up to 24. This technique works well with any number between 26 and 74. It's based on a simple algebra concept. Perhaps I'll do a video sometime... :-) (or maybe tecmath already has this method in another video?)
I had another method for xy^2 for ex 34^2, you just quare the last digit, 4 which is 16 then take 1. Next you multiply the first digit to 2 then to the last digit, 3 x 2 x 4 which is 24 and add 1 you take to 25 then continue to take 2, finally quare the first digit , 3^2 is 9, and add the 2 you take to 11. So 34^2 is 1156
for those looking for the math behind it: Say the number to be squared =A so looking for A*A. Name the nearest multiple of ten = B*10 (so it ends with a zero) So in general we get A = 10*B+C, where C is the difference between the nearest multiple of 10 and the original number; in this case C can be positive or negative. (for instance if A = 71, then B= 7 and C= 7*10-711= MINUS 1) then A squared = A*A = (10*B +C)*(10*B+C) = (10*B)*(10*B) + 2*10*B*C + C*C = 10*B * (10*B + 2C) + C*C As 10*B is the nearest multiple of 10, and as 10*B +2C = (10*B + C)+C = A+C, or the original number plus the difference, it can be rewritten as A*A= (10*B)*(A+C) + C*C
Another way of deriving the math behind this would be to use conjugates: for any number 'a', which is 'b' away from the nearest multiple of 10, we have: (a+b)(a-b) = a² - b². Then, simply add b² to both sides.
my simple trick of 32^2 is 3^2 is written as 09 and 2^2 is written as 04 now mix both i.e 0904 now 3*2 is 6 and multiply with 2 i.e 12 now add 0904 to 12 in this manner 0904 12x -------- 1024
I went for coaching there I got some tricks n I have mastered them after watching videos.... Other than that u can find some in books related to speed maths
@@Harry-bd6pb hehe I already calculate fast before watching dis vid btw Im the fastest in ma class not lying swear on ma family but with this video( I have not watched yet) I may be unstoppable
For 77² I go down from 80² : so, 6400 - 6x80 + 9 My method is from drawing squares on cm/mm paper, and from there simply "visualized". So, to go from 10 to 12 squared you need to add 2 rows of 10 on the right, as well as on top, plus add 2 square to fill the "missing piece", to get the 12 by 12 square
I am a bit late but I'd like to post my theory on squares. Let's say you know a number squared...per say 2^2, but you don't know the next number squared... in this case, 3^2. Take those numbers (2 & 3) and add them (5). Now add it to the number that you know squared. Since you know what 2^2 is, add it to the number you got in result of adding the square roots(5). You get 9...AKA 3^2. This works for any square as long you know the one before. Not too sure how I found this out but hey, I like it.
Nice video! :) but while doing these questions I found another way to get the answer. Eg 77^2, the difference is 7 so you can also do (70* 84) + (7^2) still give you the same answer!
The best way to find the square of 2 digit number than this method is showing below ?=34^2 step 1= last number is 4 ryt,,square of 4=16,enter last number ".........6" balance is 1 step 2=multiply between 3*4 and double it =12*2=24,,,,,,,,plus balance 1=25,enter 5 ".......56" balance is 2 step 3=square of 3 =9 plus balance 2 =11 enter that 11.......total is "1156"
You can also jump from one square to the next like this: n+(2n^1/2)+1 Example: 16+(2(16)^1/2)+1=16+2(4)+1=25 The reverse is also possible: n-(2n^1/2)+1
For anyone wondering why this works: N= number, d= difference (from nearest ten). n^2 = (n+d) * (n-d) + d^2. This uses the identity property to get to numbers that are easier for figuring in your head.
I usually use the FOIL method..... Basically : 32×32 can be written as : (30+2)*(30+2) now multiply Fronts , that is 30*30 ...then multiply Outers ..i.e 30*2...then multiply Inners ..that is : 2*30 ... lastly multiply lasts .i.e : 2*2 ... might look tough first but it's pretty easy once you get the hang of it...now add everything ...so 900+60+60+4 = 1024 ... Easy
Is that done using a vedic math sutra? I haven't learned that one yet but I can do 35^2 35^2 1 more than 3 is 4 so.. 3x4=12\5^2 = 12\25 = 1,225 And another: 85^2 1 more than 8 is 9 so.. 8x9=72\5^5 = 72\25 = 7,225
Thats too many steps... May as well multiply 32 and 32. If your that fast at multiplying 30 and 34 and adding the square and all that stuff,what is so hard about multiplying 32 and 32 and getting it over with?
It's way easier trying to multiply 30 and 34, since you can turn it into 3*34, which is super easy, 90 + 12 = 102, and then you add a zero, since you took away the zero from 30. That makes 1020, then you add 2^2, which makes 1024. It might seem like a lot of steps, but trust me, I got the answer in less than a few seconds. I'm younger than 11 years old. Also, I did all this mentally, so try it yourself, and hopefully, you can change your mind.
This is basic quadratic equation that you have described! 77*77 = (70+7)*(70+7) => 70*70 + 2*70*7 + 7*7 ==> Now take 70 as common factor from first two terms of this multiplication ==> 70 (70 + 2*7) + 7*7 ==> 70 * 84 + 7*7 ==> This is essentially the same result as you have described!
For 3-digits, it's the same technique it's just that you need to round that number to it's nearest ten. Ex: 273 = 270 471 = 470 298 = 300 So it's easier to add/subtract. Also that the base multiplying by the sum/difference is the same. It's just a 3 digit number.
Yes, it's called finding the Square Root. It's possible but I've never been able to do it without a calculator or computer with a Square Root function available.
H2oFormula rcmodelr Yes, there is super easy way to find square root without a calculator long as answer is a whole number NOT a decimal. here how square numbers works. Square of 1 and 9 end with 1 2 and 8 end with 4 3 and 7 end with 9 4 and 6 end with 6 5 end with 5 0 end with 0 32^2 = 1024 lets reverse it and lets pretend we don't know the square root of 1024 last digit is 4 and sq root end with 4 is 2 and 8. so last digit will be 2 or 8. now do the first number (10) square of 3 is 9 square of 4 is 16 since our first number is 10 so we use square of 3 so answer will either 32 or 38. since 10 is really close to 9 than 16 so answer will be 32. and square of 38 will be close to 16. it will be 8*8 = 64 24+24 = 48 +6.4 = 54.4 = 544 3*3 = 9 (30*30=900) 900+544 = 1444 = 38^2 1444 is close to 1600
this method is based on the formula : (x-n)(x+n)+n^2=x^2. let x=ab the number with two digits, we choose n= the distance from the number x and the nearest tens number. for example 17^2 ? the nearest tens number is 20, the distance is then n=3. 17^2=(17-3)*(17+3)+3^2=14*20+9=280+9=289
For those who seek for the formula if 3 digit number occures, for example 97*97 = 90*104 but it does not give the correct answer, the formula changes, now it is 90*104 + 7*7, 7 is the digit that you put on top of your number when you break it down, if it was 95 you would add 5*5, so the formula would look like 90*100 + 5*5.
Doesnt matter u just do 50 x 60 which is 3000, than add 5^2 cuz u went up and down by 5. 3000+ 25=3025. All 2 digits that end in 5 ends in 25. Example 75= 80x70 =5600. +25=5625
It's very easy. The number will always end in 25. So lets take 75 for example. The last two digits of the square are 25, so 75²=xx25. You get the first two numbers by multiplying 7 by the nearest higher number, So 8. 7*8 is 56, so the answer is 5625
Just a quick comment to show why this trick works! The reason that you double the number in the 1's place (such as 2 in 32) to get 30 x 34 is simple. We all learn in quadratics that (a+b)^2= a^2+2ab+b^2. So in the trick we are changing 32^2 into (30+2)^2. See this would give us 30^2+(2 x 30 x 2)+2^2. What we can then do is (by the associative property) multiple 2 x 2 = 4 inside the parenthesis. See how now we have 30^2+(30 x 4)+2^2? If we break this apart we see that we're doing (30 x 30)+(30 x 4)+(2 x 2)! You can see how (30 x 30)+(30 x 4) is the exact same as (30 x 34), as shown in the trick!
a^2 - b^2 = (a+b)(a-b) change a bit: a^2 = (a+b)(a-b) + b^2 let c = a+b let d = a - b to square 32 we need to: round 32 to the ten's: 30 take the differnce: 2 now we substitute a = 32, b = 2 we calculate c = 34, d = 30 now, cd + b^2 is our answer
Well what do you think which one is better? 30*30+4*30=900+120=1020 and then just add 2^2=4 --> 1024 Or 32*32=30*32+2*32=30*30+2*30+64=900+60+64=1024 His method seems simpler in my opinion
my fav way is (30+2)^2=30^2+2(30*2)+2^2=900+1200+4=1024 or for this specific example you can spot that 32=2^5 so 32^2=2^10 and because I do a lot of binary work I know thats 1024 off the top of my head
Another good method is 96² => 9² - 81 => 6² - 36 8136 Then multiply 9×6×2= 108 Then add a 0 at the back of the no. 1080 Then 8136 +1080 =>9216 *Only works with two-digit-numbers*
That's usually my method, in my head. I picture the numbers then go through the motions and add. With practice, it becomes quicker than whipping out your smartphone and do it on the calculator.
I must say. The teacher is a very good guide. I do not know much maths .I just watched your vedio and calculated all saqures. It is brilliant and it works well. Thanks very much. I have one request that please upload any sort of vedio which shows multiplication in a quick and easy way.
ORRRR.... you can make it simpler: Let's take a number. I'll use 32 like he did in the video. I'll call digit a = 3 and digit b = 2. Take the first digit, square it, and multiply by 100: 100(3^2) = 900 Multiply the two digits together, then by 20: 3 * 2 * 20 = 120 Square the last digit: 2 ^ 2 = 4 Add them up: 900 + 120 + 4 = 1024
Yes, it makes mathematical sense. Here: (10x + y)^2 (that's 10 times the first number because it's in the tens digit) = (10x + y)*(10x + y) = 100x^2 + 20xy + y^2 So yes, it does ;)
noname noname Thanks, it means a lot! I might actually start a channel with cool math stuff like this (plus rubik's cube videos and tutorials) soon so stay tuned ;) have a wonderful day!
77 ^2, 7^2 is 49, stick on a zero. 490, add the 49,= 539, stick on a zero, 5390 and then add the 539.. also good if you can help yourself mentally get to it. Thank you for your video. this ofc, only works for double digit numbers
Thanks for the comment! Truth is....well....I have to admit....I worked it out. I do, however, often double check my questions and answers after I make a video - mainly because like everyone I can (and quite often do) make mistakes. Funnily enough, for this particular video I had to remove it and reupload it to remove a mistake I had made - the result of not double checking!
you can do it like this: for 33 for example, you do 30^2 + 3^2 + 2*3*30 = 900 + 9 + 180 = 1089 for 54, you do 50^2 + 4^2 + 2*50*4 = 2500 + 16 + 400 = 2916 for 89, you do 80^2 + 9^2 + 2*80*9 = 6400 + 81 + 1440 = 7921
This is another way of using the difference of two squares, i.e. a^2 - b^2 = (a - b)(a + b). For the first example, we wanted to find 32^2. Notice that: 30 x 34 = (32 - 2)(32 + 2) = 32^2 - 2^2. Thus, we add the 2^2 back on to compensate for this. Thanks for the video.
If you start with 1 then add each consecutive odd number first add 3 then add 5 the add 7 and so on and so forth each time you had the next consecutive odd number each answer you get will be a perfect square. (1+3=4, 4+5=9, 9+7=16 and so on)
Another way to work it out is to do 30 x 32 = 960 and 2 x 32 = 64 add them together = 1024. You get the 30 by taking last digit off 32. And then you have to get the 2 back by multiply 32 x 2.
I have noticed, if you are like me where this is like your 5th or 6th video that you've "Studied" on his channel, this is all getting easier and easier as you go along. I came on here because when i was in Elementary school I never learned how to cross multiply i slipped under the radar all through school. Now that I'm ready to start college it hit me that I should learn how to work these sort of problems. >>> Trust me on this, If you do the examples and force yourself to learn what he is showing you how to do, it will become easier than you think.
square a number also.. x^2 + y^2 +2(xy)=(xy)^2. so 48^2 is 40^2+8^2+2(40*8)=1600+64+640=2304. works for all numbers, but for 3 digit the x value becomes, for instance, 121^2 x= 120 and y = 1. y is always the units the x is the rest, so 120 ^2 is 14400, 1^2 is 1 + 240 =1461.. easy. worked this out on a rowing machine.
Another form that turns out to be quicker for numbers with 3 or more digits in most of the cases is using a binomial squared: (a+b)2 = a2 + 2ab + b2. But the trick is that you have to try that a is a multiple of 10
I found out another way to make it, to me it's easier: (Sorry for my english: I'm italian) Make the square of the first number (you write the result in the units "column") , than multiply the first and the second number and multiply again by 2 (you write it in the tens "column") and finally make the square of the second number (you write it in the hundreds "column"). Example: 32^2= units= 2^2= 4 tens= (3x2)x2= 12 (write "2" and add the "1" to the hundreds column Hundreds= 3^2= 9 (plus 1= 10) So the result is: 1024 I hope it will be useful to you👋🏼
I've always had trouble with all types of math, but watching your videos helps a lot. Why do teachers not show you these methods in school / college? Is it that they simply don't know them?
Dr. Ashwani Kumar These "silly" methods can actually be used in practical everyday applications; it's always useful to be able to do multi-digit multiplication and squaring quickly in your head. Furthermore, just because it's a higher level of math doesn't mean it's more "serious". There is not just one branch of math that is "important", much as you may think.
The method in the video works for three digit, even four digit, numbers as well. Though it is harder to multiply some numbers (for example 463, which is the same as 460*466+9)
So this method is rearranging (x+y).(x-y) = x^2 - y^2 to get x^2 = (x+y).(x-y) + y^2, and picking a suitable y to make the calculations easier by making either (x+y) or (x-y) a multiple of 10.
For 77^2, I was afraid to write 3 (to the nearest 10) worried it was going to be a -3, so I wrote a +7. 70x84=5880+49=5929. It still works. But then I can see why it makes sense to go with the smallest number possible if we're doing this mentally.
thanks a lot cause this is not one of those videos guys where you've gotta do huge complications and all....... THIS IS BRILLIANT MAN....... TECMATH ROCKS.............
I think its faster to just break that number. For example if you have number 24. So 24 sqare u can rewrite like 24x24 and now u can break it to get 20x24 + 4x24.
For me, I just separate the numbers. So for example 57² I take 50 and 7 and multiply them each by 57. Which 50 x 50 = 2500, then 50 x 7 = 350 now add 2500 and 350 and we get 2850. Now 7 x 50 = 350, then 7 x 7 which is 49. Then we add 350 + 49 = 399. And now we fully add 2850 + 399 which then we get 3249. It sounds very complicated but it's actually easier than you think. As long as you know the multiples of the numbers by 5 and 10 then you're Gucci. It takes less than a minute to do in your head and is much easier than doing it manually once you've mastered it. I wouldn't say it's better than this, I'm just showing an alternative.
Something i realised while using this amazing method is that when he says to "find the nearest ten", you have to work back to the nearest ten - 57 works to 50 not 60. But amazing. thanks!
I watched this video when it first came out, and I have been using it ever since. What for? Nothing useful, but it's fun! And I've started doing the same method with three digit numbers now - it gets easy once you have done two digit squaring for 9 years. Genuinely, finding this video again has brought on a lot of nostalgia.
there is easy way to square large numbers in head. For example... 32 ^2 32 x32 key numbers in 32 are 3, 2, and 10 (10 is used because 32 is 2 digit number) 2 * 32 = 64 divide by 10 = 6.4 3 * 32 = 96 96 +6.4 102.4 * 10 = 1024 You can also multiple 96 with 10 and then add 64 96 * 10 = 960 + 64 = 1024 But smaller numbers are easy to remember in head
I usually do it like this:
e.g. for 32 x 34:
double one side and half the other side:
32 x 34
= 16 x 68
= 8 x 136
= 4 x 272
= 2 x 544
= 1088
It's easy to do this very quick in the head.
+A4Weissalles What if you have to multiply prime numbers ;)?
+Igor Gazela Then, obviously, it would not work
But the chance of multiplying prime numbers only by mental math is vanishingly small, I think.
But any number that is not a product of two to a power can't be halved until it reaches 1, you eventually get fractions which are much harder to deal with and end in not one
It works with any prime number
e.g.:
49 x 43
= 7 x 301
= 2107
27 x 61
= 3 x 549
= 1660 - 3
= 1647
This method is not always faster than the "common method", but in many cases it is.
+A4Weissalles He/She said prime number I think. And 49/27 is not a prime number
Hey techmath,
The method you've used might get tedious for 3 digit numbers.
I've got a much easier way:
1) For 2 digit number:
NOTE: we are dealing with 2 digit number, so we must have only one digit entries.
Let's take (32)^2.
-> Square each digit and write 'em down at both ends.
Like--> 9____4. (3^2=9, 2^2=4).
-> Now need to find the middle portion of answer, for that just multiply both digit and double the result,
That implies, (3x2)x2 = 12.
-> So for that, write 2 in blank space and carry 1 to next digit. Why? Refer the NOTE above!
That is, 924 and then add 1 to 9. So we get -> 1024.
2) For 3 digit number:
Let take (409)^2.
NOTE: we are dealing with 3 digit number, so we must have only 2 digit entries and group the number in pair of two from right.
-> Group the numbers from right (in pair of two) , so we have two groups as:
Group 1:- 4, Group 2:- 09. Now the rest of the procedure is same.
-> Follow same procedure. square the two groups and write at both ends.
Like--> 16____81. (4^2=16, 9^2=81).
-> Now need to find the middle portion of answer, for that just multiply both group and double the result,
That implies, (4x9)x2 = 72.
-> So answer is, 16 72 81 = 167281.
3) Another 3 digit number:
Let's try with a tougher number:- (825)^2.
-> Group the numbers from right (in pair of two) , so we have two groups as:
Group 1:- 8, Group 2:- 25. Now the rest of the procedure is same.
-> Follow same procedure. square the two groups and write at both ends.
Like--> 64____25.
Why? Refer NOTE for 3 digit numbers. We must have only 2 digit entries for 3 digit number. And (25)^2=625. So write 25 and
take 6 as carry.---------------------------------------> (1).
-> Now need to find the middle portion of answer, for that just multiply both group and double the result,
That implies, (8x25)x2 = 400. But write only 00 and take 4 as carry----------------------------------> (2).
Hence, Result will be:- 64_____25 --> 64 00 25, But we are yet to add the carries from equation (1) and (2).
Now from (1) add carry 6 to 00.
And from (2) add carry 4 to 64.
Final answer: 68 06 25.
P.S If anyone has any doubt, ask me! I'll try to clarify! Cheers!
how about 99*99?
Hey Lon Dong! :)
Good doubt, I was expecting this somewhere down the line! I'm glad you asked.
Follow same procedure!
-> 9^2=81
Remember, the NOTE given for 2 digit no, only one digit will be written initially.
Hence,
->Write as 1____1 (8, being taken as carry from both ends).
-> Now (9x9)x2=162.
Here lies the difference, since we got 3 digit result. Follow carefully from now.
I'll rewrite the result, this time the value to be carried in brackets, just to help you understand better!
(8) (8)
-> 1___1 (Again, Remember only one digit will come in the blank space as we dealing with 2digit number).
-> 162+8=170. Put 0 in the blank space and take 1,7 as carry.
(1)
(8) (7)
-> 1 0 1. Now, it's again simple. Just add the carry to the corresponding digit below!
-> 1+7=8.
(1)
(8)
-> 8 0 1, and 1+8=9.
-> 9 8 0 1.
P.S: You'll find it difficult in the beginning, but believe me you'll be a master in this. Practice makes it perfect.
I could do these calculations in 15-20 seconds and that's only because of practice. Just remember the
NOTEs i've mentioned.
Try for 999*999! Cheers!
i get it now, it help me to do 999*999. thank u
+Lon Dong you can also apply (1000-1)^2= (1000)^2+(1)^2-2*1000*1
in case of 4 digit number , would their be only 3 digit entry?
Without watching this, I'll just say that the way I would do 32^2 is (30 + 2)^2 = 900 + 120 + 4 = 1024.
lol smart
Sagiri Babe, what we're u doin here then? 😘 Like me ( hides behind a curtain)
Where tf did you get 900?!
30^2 because of the formula (a+b)^2 = a^2 + 2ab + b^2
Nice one
I can hear you smile
That's a weird thing to say...
Star Lou it is weirder if you can't
Great
Ummmm....
This is so eerie
I swear, this method is bloody brilliant!!! I'm squaring numbers in their hundreds! Just squared 280 in seconds in my head. After attempting to square numbers between 100 and 200, and finding that numbers such as 120 which end in a zero were easier to do by writing it as (12*10)^2=12^2 * 10^2=144*100=14,400, I found this method gets difficult after 200; and I then realised that ANY number between 100 and 1000 ending with a zero is EASY to square using tecmath's method (I actually JUST now figured out you can even for numbers above those in the 200's, as I just attempted to square 980!). I'm bloody amazed and excited man I swear ^^.
You can still use this method for numbers in their hundreds which do not end in zero, however I have found they are quite difficult to do when the difference between the base you choose and the number is greater than 20, as memorising the square of numbers above 20 is infeasible, whilst those below are pretty easy. For example, choosing a base of 200, and doing 185^2: 200*170 + 15^2 = 34000 + 225 = 34225. If you aren't sure what I did, the difference between 185 and 200 is 15, so add and minus 15 from 185 and multiply those two together, and add the square of 15. If you still aren't sure then refer to tecmaths video above. Here it is easy to square 15, but if you try a number such as 167, the distance to 200 is 33, and I don't know about any of you, but I don't know 33^2 off the top of my head!
In reflection though, squaring numbers even with a distance from the base greater than 20 seems to be a lot easier and faster than the traditional long multiplication method! I just attempted to square 167, having to square 32 doing the repeat process, and it still is a lot better than the alternative.
Thank you tecmath for giving me the ability to square ANY/MOST numbers between 1 and 1000 in seconds/a-minute!
Thanks for the comment!
Glad you liked the method.
+tecmath if number is for example 75 would nearest 10 be 70 or 80?
+tecmath if number is for example 75 would nearest 10 be 70 or 80?
+Mortimor Duncan It's the same 80-75=5 / 75-70=5 so 70*80 + 5^2= 5625. It's the same.
Pedro Miguel oh ok thanks
I like the duplex method. First, I will state a few duplex patterns and maybe you will notice a pattern. The duplex method can be used to square any number with any number of digits.
D(a) = a^2 ( duplex of a single digit )
D(ab) = 2ab ( duplex of a two digit number )
D(abc) = 2ac + b^2 ( duplex of a three digit number )
D(abcd ) = 2ad + 2bc ( duplex of a four digit number )
D(abcde) = 2ae + 2bd + c^2 ( duplex of a five digit number )
so 33^2 = 3^2 I 2*3*3 I 3^2
33^2 = 9 I 18 I 9
perform any carries that are necessary and we get 1089
Thanks! Just learned about the Duplex method through your comment, can now square numbers even up to 6 digits.
For a 5 digit case it only produces 6 digits, but requires 11. Your method is incomplete
@@tonybarfridge4369 I think you might be doing it wrong. Duplex method for 5 digits must produce at least 9 digits, assuming no carrying.
@@arcwand Any explanation should come with an example. I can't see any sense in Michael's version. But I can show how to square a number of any size using cross multiplying, and also by arcing. All methods used are variations of the squaring algorithm, but cross multiplying can use mixed numbers. The small example he showed doesn't appear to match his first outline, but is a well known vedic method. IE 137^2= 13^2/13x7x2/7^2= 169/182/49= 187/6/9 (with carryovers), =18,769 [the number of digits in the RHS separations depend on how many in the 2nd root. If it was 16^2=256, the RHS would use 2 digits as in 16, and so 56 would remain in that case. In the example it uses 7^2 or one digit]
Holly crap, I can't understand why we don't learn this at school. I just figured out at 4:44 that n^2=(n-d)(n+d)+d^2.
Thanks
Good observation!!
@@venger7357 Thank you.
In some ways you learn it at school, because what you wrote is the same like: n^2 - d^2= (n+d)(n-d) and thats a binomial formula☺
@@antoniam.h.178 I certainly didn't knew about it or learnt it that way in school.
@@antoniam.h.178 yeah that’s the way we were taught it in 8th grade
0:36 O thanks! I want to thank my parents, my teachers and especially to my faithful friend, my calculator, because without it, this would not be possible.
The reason behind this is simple actually
You can write the equation of first question like this :-
(30+2)(34-2)
Let 30 be 'a', 34 be 'b' and 2 be 'c'
Now,
(a+c)(b-c)
Now ab+cb-ac-c^2
=> ab + c(b-a) - c^2
=> ab + c(2c) - c^2. (b-a = 2c)
=> ab + 2c^2 - c^2
= ab + c^2
= 30×34 + 2^2 = 1024. (Putting the values of a,b and c)
"Simple"
I have a different method which I thought of in high school which works really well for me. Let's take 32^2 again, then you find the nearest 10 just like you -> 30^2 (=900), then let's call (for the sake of the explanation) the difference between the 2 numbers: 'd' (=2 in this case). Then to get the answer you do: 900 + (30+32)*d = 900+62*2=1024.
The great thing about this method is that if you want to calculate the square of broken numbers like 41.5^2, you can do the same thing: 1600 + (40+41.5)*1.5 = 1722.25.
Or the square of 0.913? .81 + (.9+.913)*.013 = .81 + .01813 + .005439 = .833569! Okay, maybe that was a bit too hard...
both your and his method are built upon the same very basic maths:
(a+b)^2 = a^2 + 2ab + b^2
The video lacks of that explanation though, and honestly i think you could as well just use the equation i gave
@@Player-hx1gs Not quite. The method used is actually conjugates, not square of a sum. In the video, (a+b)(a-b) = a² - b², and therefore (a+b)(a-b)+b² = a². However, Flik's method is slightly different: 'd' is defined as b-a, and multiplied with a+b.
I think it is also worth mentioning that either method works for numbers on either side of the rounding up line. Like 32^2 using the method of rounding up to the higher place of ten (40), and the difference being 8, 32-8=24. Which multiplying would be (40*20=800)+(40*4=160)=960. Adding the squared difference (8^2=64) 960+64=1024.
On the other hand, using 77 and putting it to the nearest ten of 70 at a difference of 7, the multiplier would be 84. Thus it would (70*80=5600)+(70*4=280)=5880. Adding the squared difference (7^2=49), 5880+49=5929.
((Granted this is possible, I never said it was easier than your given on how to do this. I appreciate that this is out there in the first place.))
Thanks for the help. I am spending one month studying all these skills to try to improve in my calcula to no speed. This helps! Continue the good work.
O my freaking god,
This helped me soo much..
Can't really thank you enough....
The trick can be expressed by the equation
x^2 = (x + a)(x - a) + a^2
Technically, we could choose any number "a" to add and subtract from x, then add back its square, and it would still work.
But we choose "a" as the difference to the nearest 10 so that one of the (x +/- a) terms come out to round 10's, for easy mental multiplication.
For 32 squared I would do this:
Picture it as 32 X 32
Units: 2 X 2 = 4 so you write down 4
Outer and inner: (2X3)+(2X3)=12 so you write down the 2 and carry the 1
Tens: 3 X 3 = 9 plus the carried 1 = 10 so you put down the 0 and carry the 1
Add the carried 1 from the end so you have 1024
It's the quickest way to do it in my opinion and you can do it in your head as long as you can visualise it and know your times tables solidly
For anyone that wants to learn more about this method, this is called the duplex method.
@@arcwand Thank you!
You helped me so much. I'm in 8th grade and I have to do three one minute 10 problem quizzes of my squares and roots 1-25. Thanks so much! You earned yourself a subscriber!
Bro you are 20 now
@@melroy95821* almost 22 😶🌫️
fr@@melroy958
This works. It is faster than doing it by hand most of the time. Of course, you have to be used to "storing" sub-calculations in your head while you "get the rest of it." That's the challenge for me, not so much the computations.
There are a several methods that are almost instant: they are a little more sophisticated, BUT require less mental "storage" and computation. They are especially effective if you know your perfect squares up to 24 (not that hard to memorize). If you know those, you can square any number in your head REALLY fast with very little computation--like no more than two seconds. The hardest numbers are between 71 and 74, but they are still faster than the method in the video. PLEASE NOTE, I am not trying to discredit the video. It works and this guy has a bunch of great methods for doing all sorts of fun math computations that I hadn't seen and I think are wonderful.
But for squares, there are four methods used for different ranges of numbers that are almost instant and less mentally taxing.
The quickest of them is what you could call "base 50 on 25" (NOT the same base 50 method he uses in one of his multiplication videos). So, if someone asked you to square 57 (for example), you almost instantly blurt out "thirty two forty-nine" with almost no mental computation at all (only squaring the one's digit, 7, and adding 7 to 25). That forms in your head as 32, which you affix to 49.
If asked 42 squared, you almost instantly say, "seventeen sixty-four" (slightly harder than the above number, but not much). The only thing you actually did computationally in your head is subtracted 8 from 25 (8 is the difference between 50 and 42), squared that difference, and affixed your results side-by-side. That forms in your head as 50-42 is 8. 25-8 is 17. 8 squared is 64. You now affix17 and 64. 1764. That last squared quantity will take the first two place values if it is a two digit number and three place values if it has three digits, meaning you will have to "overlap" a single digit when you affix the two quantities.
So you always work with the distance between 50 and your number and "center" the result around 25: you either subtract it from 25 or add to 25. That result will start with the thousands place. You then square the distance from 50 affix it to what you had from the first step.
Here's one more: 63 squared. It is "thirty-nine sixty-nine." To get it, you think of the difference between 50 and 63. So, 13. You add that thirteen to 25. So 38. This is actually 3800, but just think of it as 38; it's less to think about and "store" in your head. You now square 13. So 169 (you should have this memorized for these techniques). Now since the 169 is three digits long this time, you "overlap" the last digit of the 38 and the first digit of the 169. The "overlapping" is really addition. What you are doing is adding 3800 to 169. But don't even think of it that way. Only think of adding the overlapping digit: the 8 and the 1. So, you now think 39 rather than 38 and affix it to the last two digits of 169, so 69. You now have a 39 and a 69 joined to 3969. Again, don't even think of the actual big numbers or that you are adding them. You had three very simple calculations and an "affixing" or joining of the results. So you never really have to do any bigger computations, assuming you've memorized your squares up to 24. This technique works well with any number between 26 and 74. It's based on a simple algebra concept. Perhaps I'll do a video sometime... :-) (or maybe tecmath already has this method in another video?)
I love you so much right noww
this is what i was looking for
Comes home after school and watch math videos about something I'm not learning while I need to study on what I'm actually learning.
When what your learning is coordinates it's fine, infact, it's more than fine.. it's great!
Just found this site and love it! I have no math in my background and am learning as much as I can. This is so much fun!
Encouraging you to always be numerate.
I had another method for xy^2 for ex 34^2, you just quare the last digit, 4 which is 16 then take 1. Next you multiply the first digit to 2 then to the last digit, 3 x 2 x 4 which is 24 and add 1 you take to 25 then continue to take 2, finally quare the first digit , 3^2 is 9, and add the 2 you take to 11. So 34^2 is 1156
for those looking for the math behind it:
Say the number to be squared =A so looking for A*A.
Name the nearest multiple of ten = B*10 (so it ends with a zero)
So in general we get A = 10*B+C, where C is the difference between the nearest multiple of 10 and the original number; in this case C can be positive or negative. (for instance if A = 71, then B= 7 and C= 7*10-711= MINUS 1)
then A squared = A*A = (10*B +C)*(10*B+C) = (10*B)*(10*B) + 2*10*B*C + C*C = 10*B * (10*B + 2C) + C*C
As 10*B is the nearest multiple of 10, and
as 10*B +2C = (10*B + C)+C = A+C, or the original number plus the difference,
it can be rewritten as A*A= (10*B)*(A+C) + C*C
Another way of deriving the math behind this would be to use conjugates: for any number 'a', which is 'b' away from the nearest multiple of 10, we have: (a+b)(a-b) = a² - b². Then, simply add b² to both sides.
my simple trick of 32^2 is 3^2 is written as 09 and 2^2 is written as 04 now mix both i.e 0904 now 3*2 is 6 and multiply with 2 i.e 12 now add 0904 to 12 in this manner
0904
12x
--------
1024
That's a beautiful solution
Thank you tiwari
You can find lot in UA-cam... Just u need to search for it
I went for coaching there I got some tricks n I have mastered them after watching videos.... Other than that u can find some in books related to speed maths
Domingo De Leon u are always welcome
You just helped so many children with their maths tests well done :)
I guess more like helped children to flex in class 😂
@@Harry-bd6pb hehe I already calculate fast before watching dis vid btw Im the fastest in ma class not lying swear on ma family but with this video( I have not watched yet) I may be unstoppable
For 77² I go down from 80² : so, 6400 - 6x80 + 9
My method is from drawing squares on cm/mm paper, and from there simply "visualized". So, to go from 10 to 12 squared you need to add 2 rows of 10 on the right, as well as on top, plus add 2 square to fill the "missing piece", to get the 12 by 12 square
Very good, Tecmath...
I teach math so I'll have to use it with my students.
ThanksX
I am a bit late but I'd like to post my theory on squares. Let's say you know a number squared...per say 2^2, but you don't know the next number squared... in this case, 3^2. Take those numbers (2 & 3) and add them (5). Now add it to the number that you know squared. Since you know what 2^2 is, add it to the number you got in result of adding the square roots(5). You get 9...AKA 3^2. This works for any square as long you know the one before. Not too sure how I found this out but hey, I like it.
Nice video! :) but while doing these questions I found another way to get the answer. Eg 77^2, the difference is 7 so you can also do (70* 84) + (7^2) still give you the same answer!
We got the same picture! Ahhhhh! =D
concode102 Naruto!!!!! 2nd Favorite Anime Show :0
Naruto!!!!! 2nd Favorite Anime Show :0
those were really cool tricks.. it helped me a lot! make more of em. thank you so much!
The best way to find the square of 2 digit number than this method is showing below
?=34^2
step 1= last number is 4 ryt,,square of 4=16,enter last number ".........6" balance is 1
step 2=multiply between 3*4 and double it =12*2=24,,,,,,,,plus balance 1=25,enter 5 ".......56" balance is 2
step 3=square of 3 =9 plus balance 2 =11
enter that 11.......total is "1156"
I believe this is cross multiplication
+Muhammed CP ur gr8
a^2+2*a*b+b^2. take 3as a and 4 as b
shashank singh thanks shashank....
+Muhammed CP do u have a trick for 3 digits? thanx for the tip btw.
You can also jump from one square to the next like this:
n+(2n^1/2)+1
Example: 16+(2(16)^1/2)+1=16+2(4)+1=25
The reverse is also possible:
n-(2n^1/2)+1
I knew what 32^2 was in the first half a second because of binary and stuff.
Brandon Boyer isn't that 2^10?
You don't need to calculed 32^2 when you do binary . You know that right ? So why did You Say You know the answer because binery ?
Bificalera1 yes u do. U need binary in squaring
to square powers we just double the exponent so 2^5 = 32
therefore (2^5)^2 = 2^10 = 1024
similarly, to take the square root of a number, we halve the exponent.
3^4 = 81
sqrt(3^4) = (3^4)^(1/2) = 3^2 = 9
thank you very much.i was finding such tricks.keep uploading.
use (a+b)^2 or (a-b)^2 instead.. its faster this way.. For 77 it'll be (80-3)^2 = 80^2 + 3^2 - 2×80×3 = 6400+9 -480=5929
For anyone wondering why this works: N= number, d= difference (from nearest ten).
n^2 = (n+d) * (n-d) + d^2.
This uses the identity property to get to numbers that are easier for figuring in your head.
Why bother multiplying 30x34....it takes just as much work to multiply 32x32
ferbritzeo nah just multiply 3(34) then add the 0 at the right side
ferbritzeo not really
ferbritzeo not really
ferbritzeo Worked it out! 1024 in 15 seconds!
ferbritzeo yeah, I was about to post the same comment.
I usually use the FOIL method.....
Basically :
32×32 can be written as : (30+2)*(30+2) now multiply Fronts , that is 30*30 ...then multiply Outers ..i.e 30*2...then multiply Inners ..that is : 2*30 ... lastly multiply lasts
.i.e : 2*2 ... might look tough first but it's pretty easy once you get the hang of it...now add everything ...so 900+60+60+4 = 1024 ... Easy
If it is 32 * 32
You could do 30 * 2 + 30 * 2 = 120
Then you would do 30 * 30 = 900
Add them 120 + 900 = 1020
Then 2 * 2 = 4
Add them 1020 + 4 = 1024
32 x 32 we can also do
a²+2ab+b²
3²+2x3x2+2²
1024
We can find the Answer.
wut
can not do like that >
That is not correct. You would have to do:
32 using
a²+2ab+b²
where a = 30, b = 2
30²+2x30x2+2² (not 3²+2x3x2+2²)
900 + 120 + 4
=1024
I like this one. This is also the n\method I choose.
Is that done using a vedic math sutra? I haven't learned that one yet but I can do 35^2
35^2
1 more than 3 is 4 so.. 3x4=12\5^2 = 12\25 = 1,225
And another:
85^2
1 more than 8 is 9 so.. 8x9=72\5^5 = 72\25 = 7,225
Thats too many steps... May as well multiply 32 and 32. If your that fast at multiplying 30 and 34 and adding the square and all that stuff,what is so hard about multiplying 32 and 32 and getting it over with?
I guess I'm not the only one who feel this way, waste of time
It's just some comment. 😉 multiplying 34 by 3 and adding a 0 is much easier than multiplying 32 by 32
RetroPsyche I agree with you but when doing numbers like 99 or 83 it’s harder to just multiply 99x99 or 83x83 do this method can be useful
It's way easier trying to multiply 30 and 34, since you can turn it into 3*34, which is super easy, 90 + 12 = 102, and then you add a zero, since you took away the zero from 30. That makes 1020, then you add 2^2, which makes 1024. It might seem like a lot of steps, but trust me, I got the answer in less than a few seconds. I'm younger than 11 years old. Also, I did all this mentally, so try it yourself, and hopefully, you can change your mind.
This is basic quadratic equation that you have described!
77*77 = (70+7)*(70+7)
=> 70*70 + 2*70*7 + 7*7
==> Now take 70 as common factor from first two terms of this multiplication
==> 70 (70 + 2*7) + 7*7
==> 70 * 84 + 7*7
==> This is essentially the same result as you have described!
I'm sorry but if we have to do 30 x 34 in our head, we might as well just do 32 x 32 in our head
Nope. It's much more easier to multiply by number ending in zero.
+Onix Not worth the extra effort.
Absolutely... Its really funny method
I was under the assumption this was all in our head
32*3 is easy and the multiply by ten
because a^2 = a^2 -b^2 + b^2
= (a-b)(a+b) +b2
we can reorganize any number into this format with b being the small number
Thank you very much it really works. Bye guys I have a BIG EXAM tomorrow. 😉(Edexcel Exam).
For 3-digits, it's the same technique it's just that you need to round that number to it's nearest ten.
Ex: 273 = 270
471 = 470
298 = 300
So it's easier to add/subtract.
Also that the base multiplying by the sum/difference is the same. It's just a 3 digit number.
Is it possible to reverse the process and find the original number with the square?
no
Yes, it's called finding the Square Root. It's possible but I've never been able to do it without a calculator or computer with a Square Root function available.
rcmodelr I work out square roots by estimating and keep estimating until i get it
H2oFormula rcmodelr Yes, there is super easy way to find square root without a calculator long as answer is a whole number NOT a decimal. here how square numbers works.
Square of
1 and 9 end with 1
2 and 8 end with 4
3 and 7 end with 9
4 and 6 end with 6
5 end with 5
0 end with 0
32^2 = 1024
lets reverse it and lets pretend we don't know the square root of 1024
last digit is 4 and sq root end with 4 is 2 and 8. so last digit will be 2 or 8.
now do the first number (10)
square of 3 is 9
square of 4 is 16
since our first number is 10 so we use square of 3
so answer will either 32 or 38.
since 10 is really close to 9 than 16 so answer will be 32.
and square of 38 will be close to 16.
it will be 8*8 = 64
24+24 = 48 +6.4 = 54.4 = 544
3*3 = 9 (30*30=900)
900+544 = 1444 = 38^2
1444 is close to 1600
Rocki like
Nice one ahahahha! I'm remembering that one xD
there is multiple ways to calculate that, fot instance 32^2= 31*33 + 1. And formula works for every square, : x=[ (x-1)(x+1)] +1.
this was so heelpfull keep on making these vids :)
Thank you so much
〖32〗^2 = 〖30〗^2 + 2(30+32) = 1024
thanks
, 32^2= 2^10= 2^6* 2^4=64+32+64= 96
64 =1024
Mukund
this method is based on the formula : (x-n)(x+n)+n^2=x^2.
let x=ab the number with two digits, we choose n= the distance from the number x and the nearest tens number.
for example 17^2 ? the nearest tens number is 20, the distance is then n=3.
17^2=(17-3)*(17+3)+3^2=14*20+9=280+9=289
WHAT!!!
32^2 was easy because 32=2^5 32^2=2^10=1024
I know numbers up to 2^16
@@lilmarionscorner please stfu we don't care you a nerd
@@FusionXZ No ways s---er
For those who seek for the formula if 3 digit number occures, for example 97*97 = 90*104 but it does not give the correct answer, the formula changes, now it is 90*104 + 7*7, 7 is the digit that you put on top of your number when you break it down, if it was 95 you would add 5*5, so the formula would look like 90*100 + 5*5.
It feels like Mr.Bean is talking in a different style...
I've said this before and ill say it again. Small channels are always better at explaining stuff. Im subscribing because of this video
what is the nearest number of 55
It's 50
Equal or less than 54.99 rounds to 40, Equal or greater than 55 rounds to 50
Mikko Julku what has that to do with rounding to the nearest 10 multiple? 🤔 you've confused me :( .
Doesnt matter u just do 50 x 60 which is 3000, than add 5^2 cuz u went up and down by 5. 3000+ 25=3025. All 2 digits that end in 5 ends in 25. Example 75= 80x70 =5600. +25=5625
It's very easy. The number will always end in 25. So lets take 75 for example. The last two digits of the square are 25, so 75²=xx25. You get the first two numbers by multiplying 7 by the nearest higher number, So 8. 7*8 is 56, so the answer is 5625
Just a quick comment to show why this trick works! The reason that you double the number in the 1's place (such as 2 in 32) to get 30 x 34 is simple. We all learn in quadratics that (a+b)^2= a^2+2ab+b^2. So in the trick we are changing 32^2 into (30+2)^2. See this would give us 30^2+(2 x 30 x 2)+2^2. What we can then do is (by the associative property) multiple 2 x 2 = 4 inside the parenthesis. See how now we have 30^2+(30 x 4)+2^2? If we break this apart we see that we're doing (30 x 30)+(30 x 4)+(2 x 2)! You can see how
(30 x 30)+(30 x 4) is the exact same as (30 x 34), as shown in the trick!
it would be easier just to multiply the number being squared by itself
Wow you're a Genius
@@ToxicTubeAgario so doing 30x34 in your head is easier than 32x32?
i just use the algebraic identities (a-b)^2 = a^2 -2ab + b^2 so 77^2 would be (80-3)^2 = 80^2 - 2*80*3 + 3^ = 5929
Can you be my school math teacher please ?
a^2 - b^2 = (a+b)(a-b)
change a bit:
a^2 = (a+b)(a-b) + b^2
let c = a+b
let d = a - b
to square 32 we need to:
round 32 to the ten's: 30
take the differnce: 2
now we substitute a = 32, b = 2
we calculate c = 34, d = 30
now, cd + b^2 is our answer
32^2 is just 32x32...
How is doing all that work to get 2 2 digit numbers which you still have to multiply easier? :S
Well what do you think which one is better?
30*30+4*30=900+120=1020 and then just add 2^2=4 --> 1024
Or 32*32=30*32+2*32=30*30+2*30+64=900+60+64=1024
His method seems simpler in my opinion
IM AMAZIMG I DID 32^ IN FIVE SECONDS
my fav way is (30+2)^2=30^2+2(30*2)+2^2=900+1200+4=1024
or for this specific example you can spot that 32=2^5 so 32^2=2^10 and because I do a lot of binary work I know thats 1024 off the top of my head
rgqwerty63 Mliih Hhh Tyara WLd Blady
rgqwerty63 tell me what 10101111 in hex is then.
Another good method is
96²
=> 9² - 81
=> 6² - 36
8136
Then multiply 9×6×2= 108
Then add a 0 at the back of the no.
1080
Then 8136
+1080
=>9216
*Only works with two-digit-numbers*
you can just multiply 32×32
yeah I agree but this is just an example to make it simple.
I made it - 1,240 in the 5 x second challenge! ;-(
That's usually my method, in my head. I picture the numbers then go through the motions and add. With practice, it becomes quicker than whipping out your smartphone and do it on the calculator.
It's 1024
William Wheeler
How is that?
I must say. The teacher is a very good guide. I do not know much maths .I just watched your vedio and calculated all saqures. It is brilliant and it works well. Thanks very much. I have one request that please upload any sort of vedio which shows multiplication in a quick and easy way.
ORRRR.... you can make it simpler:
Let's take a number. I'll use 32 like he did in the video. I'll call digit a = 3 and digit b = 2.
Take the first digit, square it, and multiply by 100: 100(3^2) = 900
Multiply the two digits together, then by 20: 3 * 2 * 20 = 120
Square the last digit: 2 ^ 2 = 4
Add them up: 900 + 120 + 4 = 1024
+Ori Yonay but does it work for every number?
Yes, it makes mathematical sense. Here:
(10x + y)^2 (that's 10 times the first number because it's in the tens digit) = (10x + y)*(10x + y) = 100x^2 + 20xy + y^2
So yes, it does ;)
noname noname Thanks, it means a lot! I might actually start a channel with cool math stuff like this (plus rubik's cube videos and tutorials) soon so stay tuned ;)
have a wonderful day!
Thanks :) do you have any tricks to squaring 3 digits number please ? :)
MaramHattab I'll make up one and I'll try to get back to you asap..
77 ^2, 7^2 is 49, stick on a zero. 490, add the 49,= 539, stick on a zero, 5390 and then add the 539.. also good if you can help yourself mentally get to it. Thank you for your video. this ofc, only works for double digit numbers
Truth is: You wrote it down and did not calculate merely in your head.;-)
Thanks for the comment!
Truth is....well....I have to admit....I worked it out. I do, however, often double check my questions and answers after I make a video - mainly because like everyone I can (and quite often do) make mistakes. Funnily enough, for this particular video I had to remove it and reupload it to remove a mistake I had made - the result of not double checking!
you can do it like this: for 33 for example, you do 30^2 + 3^2 + 2*3*30 = 900 + 9 + 180 = 1089
for 54, you do 50^2 + 4^2 + 2*50*4 = 2500 + 16 + 400 = 2916
for 89, you do 80^2 + 9^2 + 2*80*9 = 6400 + 81 + 1440 = 7921
This is another way of using the difference of two squares, i.e. a^2 - b^2 = (a - b)(a + b).
For the first example, we wanted to find 32^2. Notice that:
30 x 34 = (32 - 2)(32 + 2) = 32^2 - 2^2. Thus, we add the 2^2 back on to compensate for this.
Thanks for the video.
u r absolutely amazing...thnx
Thank you a ton!
If you start with 1 then add each consecutive odd number first add 3 then add 5 the add 7 and so on and so forth each time you had the next consecutive odd number each answer you get will be a perfect square. (1+3=4, 4+5=9, 9+7=16 and so on)
Awesome video this method is brilliant now I can square too fast than before
Thank you for being so smart. You just made me feel a lot more intelligent by explaining this method!! Why aren't taught all this in high school!
Another way to work it out is to do 30 x 32 = 960 and 2 x 32 = 64 add them together = 1024. You get the 30 by taking last digit off 32. And then you have to get the 2 back by multiply 32 x 2.
I have noticed, if you are like me where this is like your 5th or 6th video that you've "Studied" on his channel, this is all getting easier and easier as you go along. I came on here because when i was in Elementary school I never learned how to cross multiply i slipped under the radar all through school. Now that I'm ready to start college it hit me that I should learn how to work these sort of problems. >>> Trust me on this, If you do the examples and force yourself to learn what he is showing you how to do, it will become easier than you think.
square a number also.. x^2 + y^2 +2(xy)=(xy)^2. so 48^2 is 40^2+8^2+2(40*8)=1600+64+640=2304. works for all numbers, but for 3 digit the x value becomes, for instance, 121^2 x= 120 and y = 1. y is always the units the x is the rest, so 120 ^2 is 14400, 1^2 is 1 + 240 =1461.. easy. worked this out on a rowing machine.
For 2 digit numbers I usually use the fact that a squared + 2ab + B squared = (a+B) squared, in one of these cases a being 30 and B being
Another form that turns out to be quicker for numbers with 3 or more digits in most of the cases is using a binomial squared: (a+b)2 = a2 + 2ab + b2. But the trick is that you have to try that a is a multiple of 10
I found out another way to make it, to me it's easier:
(Sorry for my english: I'm italian)
Make the square of the first number (you write the result in the units "column") , than multiply the first and the second number and multiply again by 2 (you write it in the tens "column") and finally make the square of the second number (you write it in the hundreds "column").
Example: 32^2=
units= 2^2= 4
tens= (3x2)x2= 12 (write "2" and add the "1" to the hundreds column
Hundreds= 3^2= 9 (plus 1= 10)
So the result is: 1024
I hope it will be useful to you👋🏼
What do you mean?
I've always had trouble with all types of math, but watching your videos helps a lot. Why do teachers not show you these methods in school / college? Is it that they simply don't know them?
these are silly methods good for entertainment, nothing to do with serous and higher mathematics.
Dr. Ashwani Kumar These "silly" methods can actually be used in practical everyday applications; it's always useful to be able to do multi-digit multiplication and squaring quickly in your head. Furthermore, just because it's a higher level of math doesn't mean it's more "serious". There is not just one branch of math that is "important", much as you may think.
THANK YOU YOU ARE A LIFE SAVER !!!
The method in the video works for three digit, even four digit, numbers as well. Though it is harder to multiply some numbers (for example 463, which is the same as 460*466+9)
So this method is rearranging (x+y).(x-y) = x^2 - y^2 to get x^2 = (x+y).(x-y) + y^2, and picking a suitable y to make the calculations easier by making either (x+y) or (x-y) a multiple of 10.
Wow thanks for helping me I really appreciate it you made squaring easier for my test
For 77^2, I was afraid to write 3 (to the nearest 10) worried it was going to be a -3, so I wrote a +7. 70x84=5880+49=5929. It still works. But then I can see why it makes sense to go with the smallest number possible if we're doing this mentally.
thanks a lot cause this is not one of those videos guys where you've gotta do huge complications and all....... THIS IS BRILLIANT MAN....... TECMATH ROCKS.............
I have a maths test coming up and this seriously is helping me revise!!!!!! Thank you so much!!!
I think its faster to just break that number. For example if you have number 24. So 24 sqare u can rewrite like 24x24 and now u can break it to get 20x24 + 4x24.
I think it's slightly faster to just use,
(A + B)^2 = A^2 + 2AB + B^2 rather than the difference of squares eg.
33^2 = 30^2 + 2 * 30 * 3 + 3^2 = 900 + 180 + 9 = 1089.
thnks so much
neat little trick, really helpful
Thank you so much, this trick is making exponents ALOT easier.
For me, I just separate the numbers. So for example 57² I take 50 and 7 and multiply them each by 57. Which 50 x 50 = 2500, then 50 x 7 = 350 now add 2500 and 350 and we get 2850. Now 7 x 50 = 350, then 7 x 7 which is 49. Then we add 350 + 49 = 399. And now we fully add 2850 + 399 which then we get 3249. It sounds very complicated but it's actually easier than you think. As long as you know the multiples of the numbers by 5 and 10 then you're Gucci. It takes less than a minute to do in your head and is much easier than doing it manually once you've mastered it. I wouldn't say it's better than this, I'm just showing an alternative.
Currently studying the ASVAB Arithmetic Reasoning subtest. And one of these questions definitely messed me up. This helps out a lot!
Wow this is amazing I didn't think it would be that easy
Something i realised while using this amazing method is that when he says to "find the nearest ten", you have to work back to the nearest ten - 57 works to 50 not 60.
But amazing. thanks!
I watched this video when it first came out, and I have been using it ever since. What for? Nothing useful, but it's fun! And I've started doing the same method with three digit numbers now - it gets easy once you have done two digit squaring for 9 years.
Genuinely, finding this video again has brought on a lot of nostalgia.
that'd be cool if you had downloadable quizes in the description box to really grind in this method.
Other than that, ur awesome. thank you so much!
there is easy way to square large numbers in head. For example...
32 ^2
32
x32
key numbers in 32 are 3, 2, and 10 (10 is used because 32 is 2 digit number)
2 * 32 = 64 divide by 10 = 6.4
3 * 32 = 96
96
+6.4
102.4 * 10 = 1024
You can also multiple 96 with 10 and then add 64 96 * 10 = 960 + 64 = 1024
But smaller numbers are easy to remember in head
Thank you!